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受欢迎的微积分 >

integral of 1/(2xsqrt(16x^2-8))

解答

\int \frac{1}{2 x 16 x ^{2} - 8} d x

解答

\frac{1}{4 2} arcsec (2 x) + C

求解步骤

\int \frac{1}{2 x 16 x ^{2} - 8} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{x 16 x ^{2} - 8} d x

使用三角换元法

= \frac{1}{2} \cdot \int \frac{1}{2 2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2 2} \cdot \int 1 d u

常数积分:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{2 2} \cdot 1 \cdot u

u = arcsec (\frac{4}{2 2} x)

代回

= \frac{1}{2} \cdot \frac{1}{2 2} \cdot 1 \cdot arcsec (\frac{4}{2 2} x)

化简

\frac{1}{2} \cdot \frac{1}{2 2} \cdot 1 \cdot arcsec (\frac{4}{2 2} x) : \frac{1}{4 2} arcsec (2 x)

= \frac{1}{4 2} arcsec (2 x)

解答补常数

= \frac{1}{4 2} arcsec (2 x) + C

作图

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